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\begin{align} -20 &= -20\\ 16-36 &= 25-45\\ 4^2-4\times 9&=5^2-5\times 9\\ 4^2-4\times 9+81/4&=5^2-5\times 9+81/4\\ 4^2-4\times 9+(9/2)^2&=5^2-5\times 9+(9/2)^2\\ \end{align}

Considering the formula $a^2+2ab+b^2=(a-b)^2$, one has \begin{align} (4-9/2)^2&=(5-9/2)^2\\ \sqrt{(4-9/2)^2}&=\sqrt{(5-9/2)^2}\\ 4-9/2&=5-9/2\\ 4&=5\\ 4-4&=5-4\\ 0&=1 \end{align}

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closed as off-topic by Elliot G, BruceET, Claude Leibovici, user91500, Ian Miller Dec 31 '16 at 11:10

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  • 3
    $\begingroup$ $\sqrt{x^2}=|x|$ so $\sqrt{a^2}=\sqrt{b^2}$ does not imply $a=b$ but rather $|a|=|b|$. And, indeed, $|4-9/2|=|5-9/2|=1/2$ but $4-9/2 \ne 5-9/2$ and the rest doesn't follow. $\endgroup$ – dxiv Dec 31 '16 at 3:44
  • $\begingroup$ I don't see why this question keeps getting downvotes. I'll give it a +1 just because it is very clearly titled "spot the mistake", it is written down in full detail, and correctly tagged as fake-proofs. $\endgroup$ – dxiv Dec 31 '16 at 4:12
  • 1
    $\begingroup$ Possible duplicate of Best Fake Proofs? (A M.SE April Fools Day collection) $\endgroup$ – Moo Dec 31 '16 at 4:16
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Here let me simplify your process:

$$ \begin{align} (-2)^2 = 4 &\implies \sqrt{(-2)^2} = \sqrt{2^2} \\ &\implies -2 = 2 \\ &\implies -2 + 2 = 2 +2 \\ &\implies 0 = 4 \end{align} $$

QED. Do you see the mistake?

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  • $\begingroup$ So, I'm simply give the answer: we cannot root the negative integer,as in question &radic;(4-9/2)<sup>2</sup>=&radic;(5-9/2)<sup>2</sup> explaination: LHS:4-9/2=(8-9)/2 so &radic;((8-9)/2)<sup>2</sup> &radic;(-1/2)<sup>2</sup> &radic;(1/4) (1/2) RHS:5-9/2=(10-9)/2 so &radic;((10-9)/2)<sup>2</sup> &radic;(1/2)<sup>2</sup> &radic;(1/4) (1/2) LHS=RHS $\endgroup$ – Animesh Sahu Dec 31 '16 at 17:01
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$\sqrt {a^2}\ne a $. You should never have been taught that it does.

Instead $\sqrt {a^2}=|a| $ .

So....

$(4-9/2)^2=(5-9/2)^2$

$\sqrt {(4-9/2)^2}=\sqrt {(5-9/2)^2}$

$|4-9/2|=|5-9/2|$

$4 - 9/2 = \pm 5 \mp 9/2$

$4 = 5$ or $4 = -5 +2* 9/2=-5+9$

$4-4=5-4$ or $4-4 =-5-4+9$

$0=1$ or $0 = 0$

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  • $\begingroup$ So, I'm simply give the answer: we cannot root the negative integer,as in question &radic;(4-9/2)<sup>2</sup>=&radic;(5-9/2)<sup>2</sup> explaination: LHS:4-9/2=(8-9)/2 so &radic;((8-9)/2)<sup>2</sup> &radic;(-1/2)<sup>2</sup> &radic;(1/4) (1/2) RHS:5-9/2=(10-9)/2 so &radic;((10-9)/2)<sup>2</sup> &radic;(1/2)<sup>2</sup> &radic;(1/4) (1/2) LHS=RHS $\endgroup$ – Animesh Sahu Dec 31 '16 at 17:00
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    $\begingroup$ Not quite. $(5-9/2)^2 = 1/4$ and $(4-9/2)^2 = 1/4$ and that's fine. But if $a^2 = b^2$ that does NOT mean $a = b$. The problem $x^2 = 25$ has two answers. $x = 5$ is one answer but $x = -5$ is another. If $c > 0$ then there is one positive number $r$ so that $r^2 = c$ and there is one negative number that is equal to $-r$ so that $(-r)^2 = c$. So $a^2 = b^2$ does not mean $a = b$ but it means $|a| = |b|$ but either a or b could be positive or negative. So you have $(4-9/2)^2 = (5-9/2)^2$ which is fine But 4 - 9/2 < 0 while 5 - 9/2 > 0. so... to be continued... $\endgroup$ – fleablood Dec 31 '16 at 17:36
  • $\begingroup$ ... continued.... and you have $(4-9/2)^2 = (5-9/2)^2$ which is fine (they both equal $1/4$). And you have $\sqrt{(4- 9/2)^2} = \sqrt{5-9/2)^2}$ which is also fine (they both equal $1/2$). But you have $\sqrt{(4-9/2)^2} = 4-9/2 = - 1/2$ which is NOT fine. This isn't true. $\sqrt{(-5)^2} = \sqrt{25} = 5 \ne -5$. $\sqrt{a^2} \ne a$--- you should never have learned that. $\sqrt{a^2} = |a|$. $|a| = 0$ if $a \ge 0$ but $|a| = -a$ if $a \le 0$. So you can conclude $|4- 9/2| = |5-9/2|$ which is fine. They both equal 1/2. $\endgroup$ – fleablood Dec 31 '16 at 17:43

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